/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QReductionProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QReductionProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPOrderProof [EQUIVALENT, 16 ms]
        (20) QDP
        (21) PisEmptyProof [EQUIVALENT, 0 ms]
        (22) YES


----------------------------------------

(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   cond(true, x) -> cond(odd(x), p(p(p(x))))
   odd(0) -> false
   odd(s(0)) -> true
   odd(s(s(x))) -> odd(x)
   p(0) -> 0
   p(s(x)) -> x

Q is empty.

----------------------------------------

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is 
   odd(0) -> false
   odd(s(0)) -> true
   odd(s(s(x))) -> odd(x)
   p(0) -> 0
   p(s(x)) -> x

The TRS R 2 is 
   cond(true, x) -> cond(odd(x), p(p(p(x))))

The signature Sigma is {cond_2}
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   cond(true, x) -> cond(odd(x), p(p(p(x))))
   odd(0) -> false
   odd(s(0)) -> true
   odd(s(s(x))) -> odd(x)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond(true, x0)
   odd(0)
   odd(s(0))
   odd(s(s(x0)))
   p(0)
   p(s(x0))


----------------------------------------

(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND(true, x) -> COND(odd(x), p(p(p(x))))
   COND(true, x) -> ODD(x)
   COND(true, x) -> P(p(p(x)))
   COND(true, x) -> P(p(x))
   COND(true, x) -> P(x)
   ODD(s(s(x))) -> ODD(x)

The TRS R consists of the following rules:

   cond(true, x) -> cond(odd(x), p(p(p(x))))
   odd(0) -> false
   odd(s(0)) -> true
   odd(s(s(x))) -> odd(x)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond(true, x0)
   odd(0)
   odd(s(0))
   odd(s(s(x0)))
   p(0)
   p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
----------------------------------------

(6)
Complex Obligation (AND)

----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ODD(s(s(x))) -> ODD(x)

The TRS R consists of the following rules:

   cond(true, x) -> cond(odd(x), p(p(p(x))))
   odd(0) -> false
   odd(s(0)) -> true
   odd(s(s(x))) -> odd(x)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond(true, x0)
   odd(0)
   odd(s(0))
   odd(s(s(x0)))
   p(0)
   p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(8) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ODD(s(s(x))) -> ODD(x)

R is empty.
The set Q consists of the following terms:

   cond(true, x0)
   odd(0)
   odd(s(0))
   odd(s(s(x0)))
   p(0)
   p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(10) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cond(true, x0)
   odd(0)
   odd(s(0))
   odd(s(s(x0)))
   p(0)
   p(s(x0))


----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   ODD(s(s(x))) -> ODD(x)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(12) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*ODD(s(s(x))) -> ODD(x)
The graph contains the following edges 1 > 1


----------------------------------------

(13)
YES

----------------------------------------

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND(true, x) -> COND(odd(x), p(p(p(x))))

The TRS R consists of the following rules:

   cond(true, x) -> cond(odd(x), p(p(p(x))))
   odd(0) -> false
   odd(s(0)) -> true
   odd(s(s(x))) -> odd(x)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond(true, x0)
   odd(0)
   odd(s(0))
   odd(s(s(x0)))
   p(0)
   p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(15) UsableRulesProof (EQUIVALENT)
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND(true, x) -> COND(odd(x), p(p(p(x))))

The TRS R consists of the following rules:

   odd(0) -> false
   odd(s(0)) -> true
   odd(s(s(x))) -> odd(x)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   cond(true, x0)
   odd(0)
   odd(s(0))
   odd(s(s(x0)))
   p(0)
   p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(17) QReductionProof (EQUIVALENT)
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

   cond(true, x0)


----------------------------------------

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   COND(true, x) -> COND(odd(x), p(p(p(x))))

The TRS R consists of the following rules:

   odd(0) -> false
   odd(s(0)) -> true
   odd(s(s(x))) -> odd(x)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   odd(0)
   odd(s(0))
   odd(s(s(x0)))
   p(0)
   p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(19) QDPOrderProof (EQUIVALENT)
We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.

   COND(true, x) -> COND(odd(x), p(p(p(x))))
The remaining pairs can at least be oriented weakly.
Used ordering:  Polynomial interpretation [POLO,RATPOLO]:

   POL(0) = 0
   POL(COND(x_1, x_2)) = [1/4]x_1 + [1/4]x_2
   POL(false) = 0
   POL(odd(x_1)) = [1/4]x_1
   POL(p(x_1)) = [1/4]x_1
   POL(s(x_1)) = [1] + [4]x_1
   POL(true) = [1/4]
The value of delta used in the strict ordering is 1/16.
The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

   odd(0) -> false
   odd(s(0)) -> true
   odd(s(s(x))) -> odd(x)
   p(0) -> 0
   p(s(x)) -> x


----------------------------------------

(20)
Obligation:
Q DP problem:
P is empty.
The TRS R consists of the following rules:

   odd(0) -> false
   odd(s(0)) -> true
   odd(s(s(x))) -> odd(x)
   p(0) -> 0
   p(s(x)) -> x

The set Q consists of the following terms:

   odd(0)
   odd(s(0))
   odd(s(s(x0)))
   p(0)
   p(s(x0))

We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(21) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(22)
YES